I'm reading Sheldon Axler's Measure, Integration & Real Analysis and I am confused about a step that the reader is asked to verify. The theorem is:
Suppose $\left\{e_k\right\}_{k \in \Gamma}$ is an orthonormal family in a Hilbert space $V$. Then $\overline{\text { span }\left\{e_k\right\}_{k \in \Gamma}}=\left\{\sum_{k \in \Gamma} \alpha_k e_k:\left\{\alpha_k\right\}_{k \in \Gamma}\right.$ is a family in $\mathrm{F}$ and $\left.\sum_{k \in \Gamma}\left|\alpha_k\right|^2<\infty\right\}$.
Then after proving $\sum_{k \in \Gamma} \alpha_k e_k \in \overline{\operatorname{span}\left\{e_k\right\}_{k \in \Gamma}}$ the inclusion in the other direction starts:
To prove the inclusion in the other direction, now suppose $f \in \overline{\operatorname{span}\left\{e_k\right\}_{k \in \Gamma}}$. Let $$ (8.59) \;\;\;\;\;g=\sum_{k \in \Gamma}\left\langle f, e_k\right\rangle e_k, $$ where the sum above converges by Bessel's inequality (8.57) and by $8.54(\mathrm{a})$. The direction of the inclusion that we just proved implies that $g \in \overline{\operatorname{span}\left\{e_k\right\}_{k \in \Gamma}}$. Thus $$ (8.60) \;\;\;\;\;g-f \in \overline{\operatorname{span}\left\{e_k\right\}_{k \in \Gamma}} . $$ Equation 8.59 implies that $\left\langle g, e_j\right\rangle=\left\langle f, e_j\right\rangle$ for each $j \in \Gamma$, as you should verify (which will require using the Cauchy-Schwarz inequality if done rigorously).
I know that this verification can be shown by first proving the continuity of the inner product for unordered sums (which I assume requires using Cauchy Schwartz) however, I cant help but feel that there is a really obvious way of doing this that I am missing. So, if it exists, I would appreciate any help in pointing it out to me.
Let $j \in \Gamma$. Let $\varepsilon \in (0, \infty )$. By the definition of convergence of an unordered sum, there exists a finite subset $F$ of $\Gamma$ with $j\in F$ such that
$$\left\| g - \sum_{k\in F}\langle f,e_{k}\rangle e_{k} \right\| < \varepsilon .$$
It can be directly verified that as $j\in F$ and as $\{e_{k} : k\in\Gamma \}$ is an orthonormal family,
$$\left\langle \sum_{k\in F}\langle f,e_{k}\rangle e_{k}, e_{j} \right\rangle = \langle f, e_{j} \rangle .$$
By the Cauchy-Schwarz inequality,
\begin{align*} |\langle g,e_{j}\rangle - \langle f, e_{j}\rangle | &= \left|\left\langle g - \sum_{k\in F}\langle f,e_{k}\rangle e_{k}, e_{j} \right\rangle\right| \\ &\leq \left\| g - \sum_{k\in F}\langle f,e_{k}\rangle e_{k} \right\| \cdot \left\|e_{j}\right\| \\ & < \varepsilon . \end{align*}
The result follows.